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I am reading a proof from a paper I found online, and it goes like this: We want to show that there exists a surjection $f$ from the cantor set $\mathfrak{C}$ to the interval $[0,1]$ then we can show that $|\mathfrak{C}|\geqslant [0,1]$, but since $\mathfrak{C}\subseteq [0,1]$ it follows that $|[0,1]|=|\mathfrak{C}|$. Since the unit interval is uncountable, the Cantor set must be uncountable as well.

I am stuck on one word here:

After showing that for each $p\in \mathfrak{C}$ none of its digits can equal 1, we then define a function $f : C \to [0, 1]$ by taking the number not consisting entirely of the digits $\{0, 2\}$ (why not entirely?) and replacing each occurrence of $\{2\}$ by $\{1\}$ in its representation. To show that $f$ is surjective, consider any element in $a\in[0, 1]$. Represent $a$ in its binary form, and then replace each occurrence of $\{1\}$ digit by a $\{2\}$. This new number, which we label $b$, satisies $f(b) = a$. As a result, $f$ is surjective. And then it follows that follows that $|[0,1]|=|\mathfrak{C}|$.

I can't seem to see the necessity of the word entirely I can't see why the numbers $1=0.2222...$ and $0=0.000...$ would hurt anything, and $0,1\in\mathfrak{C}$, so if $f$ doesn't consider those numbers then it is an inconsistent mapping from $\mathfrak{C}$ to $[0,1]$. (by inconsistent I mean that $\text{Dom}f\neq \mathfrak{C}$).

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I also fail to see why the author would want to say "entirely" there. Also it seems to break the proof, because the domain of the function $f$ would no longer be the entire Cantor set (as you point out in the last paragraph, although I wouldn't use the word "inconsistent.") – Trevor Wilson Oct 26 '13 at 2:25
@TrevorWilson There is a book Fondations of Abstract Analysis: Second Edition by Jewgeni H. Dshalalow, I got the term inconsistent function from there. He distinguishes be between a mapping $f:A\to B$, if $\text{Dom}f=A$ then $f$ is consistent, and if this is not the case then $f$ is not consistent. Later he discusses how we can make $f$ consistent by defining $f$ from the power set of $A$ to the power set of $B$. I haven't got to a part when this was useful yet (I only use it as a supplement to the book we use in our course). But, I like the authors precision, its worth checking out. – JimmyJackson Oct 26 '13 at 3:48 me the notation $f:A \to B$ means that $A$ is the domain of $f$. But maybe analysts do things differently. – Trevor Wilson Oct 26 '13 at 5:38
up vote 1 down vote accepted

I think what “taking the number not consisting entirely of the digits $\{0,2\}$ and replacing each occurrence of $\{2\}$ by $\{1\}$ in its representation” means it the following. Take any number $b\in\mathfrak C$. Then, the ternary representation of $b$ consists entirely of the digits $\{0,2\}$ (i.e., it does not contain any digit $1$). Let $a$ be the number constructed from $b$ in such a way that each digit $2$ in $b$ is replaced by $1$ in $a$ (and you leave the zeros alone). Define $f(b)=a$.

Then, take any $a\in[0,1]$. The aim to show that $a=f(b)$ for some $b\in\mathfrak C$. Consider the binary representation of $a$. Replace all digits $1$ in the binary representation in $a$ by $2$ and view this new number as a ternary representation. Call this new number $b$. Since $b$ consists entirely of the digits $\{0,2\}$, $b\in\mathfrak C$, and $f(b)=a$ by the construction of $f$. Hence, $f$ is surjective.

Upon a closer look, I might have misunderstood the source of the confusion (metaconfusion :-)). You may just need to replace “not consisting” by “consisting.” That is, you just consider those numbers whose ternary representations consist entirely of $\{0,2\}$ The set of such numbers is $\mathfrak C$, issues with uniqueness of representations aside.

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Thanks! That makes sense maybe the author accidentally placed the 'not' in front the consisting. Its amazing how one little word can have such a drastic effect on a proof, but I guess that what makes math so fun; nothing should be ambiguous. – JimmyJackson Oct 26 '13 at 3:34
@JimmyJackson I know, right? Even the most authoritative sources are full of typos (which is natural), but I usually get as confused as you did now when I read something that is obviously not true! I think we just sometimes need to rely on our common sense and not be afraid of mentally correcting the author when something looks really off. – triple_sec Oct 26 '13 at 3:38
I agree! But, I am not so confident in my mathematics. With time I will hopefully be able to make those corrections on my own: until then I am thankful to the math community here for all of its guidance. – JimmyJackson Oct 26 '13 at 3:52

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